\(\int (e x)^{-1+n} (a+b \csc (c+d x^n))^2 \, dx\) [76]

Optimal result

Integrand size = 22, antiderivative size = 80 \[ \int (e x)^{-1+n} \left (a+b \csc \left (c+d x^n\right )\right )^2 \, dx=\frac {a^2 (e x)^n}{e n}-\frac {2 a b x^{-n} (e x)^n \text {arctanh}\left (\cos \left (c+d x^n\right )\right )}{d e n}-\frac {b^2 x^{-n} (e x)^n \cot \left (c+d x^n\right )}{d e n} \] Output:

a^2*(e*x)^n/e/n-2*a*b*(e*x)^n*arctanh(cos(c+d*x^n))/d/e/n/(x^n)-b^2*(e*x)^ 
n*cot(c+d*x^n)/d/e/n/(x^n)
 

Mathematica [A] (verified)

Time = 1.36 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.28 \[ \int (e x)^{-1+n} \left (a+b \csc \left (c+d x^n\right )\right )^2 \, dx=\frac {x^{-n} (e x)^n \left (-b^2 \cot \left (\frac {1}{2} \left (c+d x^n\right )\right )+2 a \left (a c+a d x^n-2 b \log \left (\cos \left (\frac {1}{2} \left (c+d x^n\right )\right )\right )+2 b \log \left (\sin \left (\frac {1}{2} \left (c+d x^n\right )\right )\right )\right )+b^2 \tan \left (\frac {1}{2} \left (c+d x^n\right )\right )\right )}{2 d e n} \] Input:

Integrate[(e*x)^(-1 + n)*(a + b*Csc[c + d*x^n])^2,x]
 

Output:

((e*x)^n*(-(b^2*Cot[(c + d*x^n)/2]) + 2*a*(a*c + a*d*x^n - 2*b*Log[Cos[(c 
+ d*x^n)/2]] + 2*b*Log[Sin[(c + d*x^n)/2]]) + b^2*Tan[(c + d*x^n)/2]))/(2* 
d*e*n*x^n)
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.71, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {4697, 4693, 3042, 4260, 3042, 4254, 24, 4257}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int (e x)^{n-1} \left (a+b \csc \left (c+d x^n\right )\right )^2 \, dx\)

\(\Big \downarrow \) 4697

\(\displaystyle \frac {x^{-n} (e x)^n \int x^{n-1} \left (a+b \csc \left (d x^n+c\right )\right )^2dx}{e}\)

\(\Big \downarrow \) 4693

\(\displaystyle \frac {x^{-n} (e x)^n \int \left (a+b \csc \left (d x^n+c\right )\right )^2dx^n}{e n}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {x^{-n} (e x)^n \int \left (a+b \csc \left (d x^n+c\right )\right )^2dx^n}{e n}\)

\(\Big \downarrow \) 4260

\(\displaystyle \frac {x^{-n} (e x)^n \left (2 a b \int \csc \left (d x^n+c\right )dx^n+b^2 \int \csc ^2\left (d x^n+c\right )dx^n+a^2 x^n\right )}{e n}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {x^{-n} (e x)^n \left (2 a b \int \csc \left (d x^n+c\right )dx^n+b^2 \int \csc \left (d x^n+c\right )^2dx^n+a^2 x^n\right )}{e n}\)

\(\Big \downarrow \) 4254

\(\displaystyle \frac {x^{-n} (e x)^n \left (2 a b \int \csc \left (d x^n+c\right )dx^n-\frac {b^2 \int 1d\cot \left (d x^n+c\right )}{d}+a^2 x^n\right )}{e n}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {x^{-n} (e x)^n \left (2 a b \int \csc \left (d x^n+c\right )dx^n+a^2 x^n-\frac {b^2 \cot \left (c+d x^n\right )}{d}\right )}{e n}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {x^{-n} (e x)^n \left (a^2 x^n-\frac {2 a b \text {arctanh}\left (\cos \left (c+d x^n\right )\right )}{d}-\frac {b^2 \cot \left (c+d x^n\right )}{d}\right )}{e n}\)

Input:

Int[(e*x)^(-1 + n)*(a + b*Csc[c + d*x^n])^2,x]
 

Output:

((e*x)^n*(a^2*x^n - (2*a*b*ArcTanh[Cos[c + d*x^n]])/d - (b^2*Cot[c + d*x^n 
])/d))/(e*n*x^n)
 

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4254

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1)   Subst[Int[Exp 
andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, 
 d}, x] && IGtQ[n/2, 0]
 

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

rule 4260

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Simp[a^2*x, x] + 
 (Simp[2*a*b   Int[Csc[c + d*x], x], x] + Simp[b^2   Int[Csc[c + d*x]^2, x] 
, x]) /; FreeQ[{a, b, c, d}, x]
 

rule 4693

Int[((a_.) + Csc[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol 
] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Csc[c + d*x])^ 
p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[(m + 
 1)/n], 0] && IntegerQ[p]
 

rule 4697

Int[((a_.) + Csc[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*((e_)*(x_))^(m_.), x 
_Symbol] :> Simp[e^IntPart[m]*((e*x)^FracPart[m]/x^FracPart[m])   Int[x^m*( 
a + b*Csc[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.64 (sec) , antiderivative size = 275, normalized size of antiderivative = 3.44

method	result	size
risch	\(\frac {a^{2} x \,{\mathrm e}^{\frac {\left (-1+n \right ) \left (-i \operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right ) \pi +i \operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i e x \right )^{2} \pi +i \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )^{2} \pi -i \operatorname {csgn}\left (i e x \right )^{3} \pi +2 \ln \left (x \right )+2 \ln \left (e \right )\right )}{2}}}{n}-\frac {2 i x \,b^{2} {\mathrm e}^{\frac {\left (-1+n \right ) \left (-i \operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right ) \pi +i \operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i e x \right )^{2} \pi +i \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )^{2} \pi -i \operatorname {csgn}\left (i e x \right )^{3} \pi +2 \ln \left (x \right )+2 \ln \left (e \right )\right )}{2}} x^{-n}}{d n \left ({\mathrm e}^{2 i \left (c +d \,x^{n}\right )}-1\right )}-\frac {4 \,\operatorname {arctanh}\left ({\mathrm e}^{i \left (c +d \,x^{n}\right )}\right ) e^{n} b a \,{\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i e x \right ) \left (-1+n \right ) \left (\operatorname {csgn}\left (i e x \right )-\operatorname {csgn}\left (i x \right )\right ) \left (-\operatorname {csgn}\left (i e x \right )+\operatorname {csgn}\left (i e \right )\right )}{2}}}{d e n}\)	\(275\)

Input:

int((e*x)^(-1+n)*(a+b*csc(c+d*x^n))^2,x,method=_RETURNVERBOSE)
 

Output:

a^2/n*x*exp(1/2*(-1+n)*(-I*csgn(I*e)*csgn(I*x)*csgn(I*e*x)*Pi+I*csgn(I*e)* 
csgn(I*e*x)^2*Pi+I*csgn(I*x)*csgn(I*e*x)^2*Pi-I*csgn(I*e*x)^3*Pi+2*ln(x)+2 
*ln(e)))-2*I*x*b^2*exp(1/2*(-1+n)*(-I*csgn(I*e)*csgn(I*x)*csgn(I*e*x)*Pi+I 
*csgn(I*e)*csgn(I*e*x)^2*Pi+I*csgn(I*x)*csgn(I*e*x)^2*Pi-I*csgn(I*e*x)^3*P 
i+2*ln(x)+2*ln(e)))/d/n/(x^n)/(exp(2*I*(c+d*x^n))-1)-4*arctanh(exp(I*(c+d* 
x^n)))/d/e*e^n/n*b*a*exp(1/2*I*Pi*csgn(I*e*x)*(-1+n)*(csgn(I*e*x)-csgn(I*x 
))*(-csgn(I*e*x)+csgn(I*e)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.45 \[ \int (e x)^{-1+n} \left (a+b \csc \left (c+d x^n\right )\right )^2 \, dx=\frac {a^{2} d e^{n - 1} x^{n} \sin \left (d x^{n} + c\right ) - a b e^{n - 1} \log \left (\frac {1}{2} \, \cos \left (d x^{n} + c\right ) + \frac {1}{2}\right ) \sin \left (d x^{n} + c\right ) + a b e^{n - 1} \log \left (-\frac {1}{2} \, \cos \left (d x^{n} + c\right ) + \frac {1}{2}\right ) \sin \left (d x^{n} + c\right ) - b^{2} e^{n - 1} \cos \left (d x^{n} + c\right )}{d n \sin \left (d x^{n} + c\right )} \] Input:

integrate((e*x)^(-1+n)*(a+b*csc(c+d*x^n))^2,x, algorithm="fricas")
 

Output:

(a^2*d*e^(n - 1)*x^n*sin(d*x^n + c) - a*b*e^(n - 1)*log(1/2*cos(d*x^n + c) 
 + 1/2)*sin(d*x^n + c) + a*b*e^(n - 1)*log(-1/2*cos(d*x^n + c) + 1/2)*sin( 
d*x^n + c) - b^2*e^(n - 1)*cos(d*x^n + c))/(d*n*sin(d*x^n + c))
 

Sympy [F]

\[ \int (e x)^{-1+n} \left (a+b \csc \left (c+d x^n\right )\right )^2 \, dx=\int \left (e x\right )^{n - 1} \left (a + b \csc {\left (c + d x^{n} \right )}\right )^{2}\, dx \] Input:

integrate((e*x)**(-1+n)*(a+b*csc(c+d*x**n))**2,x)
 

Output:

Integral((e*x)**(n - 1)*(a + b*csc(c + d*x**n))**2, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 207 vs. \(2 (80) = 160\).

Time = 0.07 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.59 \[ \int (e x)^{-1+n} \left (a+b \csc \left (c+d x^n\right )\right )^2 \, dx=-\frac {2 \, b^{2} e^{n} \sin \left (2 \, d x^{n} + 2 \, c\right )}{d e n \cos \left (2 \, d x^{n} + 2 \, c\right )^{2} + d e n \sin \left (2 \, d x^{n} + 2 \, c\right )^{2} - 2 \, d e n \cos \left (2 \, d x^{n} + 2 \, c\right ) + d e n} + \frac {\left (e x\right )^{n} a^{2}}{e n} - \frac {{\left (e^{n} \log \left (\cos \left (d x^{n}\right )^{2} + 2 \, \cos \left (d x^{n}\right ) \cos \left (c\right ) + \cos \left (c\right )^{2} + \sin \left (d x^{n}\right )^{2} - 2 \, \sin \left (d x^{n}\right ) \sin \left (c\right ) + \sin \left (c\right )^{2}\right ) - e^{n} \log \left (\cos \left (d x^{n}\right )^{2} - 2 \, \cos \left (d x^{n}\right ) \cos \left (c\right ) + \cos \left (c\right )^{2} + \sin \left (d x^{n}\right )^{2} + 2 \, \sin \left (d x^{n}\right ) \sin \left (c\right ) + \sin \left (c\right )^{2}\right )\right )} a b}{d e n} \] Input:

integrate((e*x)^(-1+n)*(a+b*csc(c+d*x^n))^2,x, algorithm="maxima")
 

Output:

-2*b^2*e^n*sin(2*d*x^n + 2*c)/(d*e*n*cos(2*d*x^n + 2*c)^2 + d*e*n*sin(2*d* 
x^n + 2*c)^2 - 2*d*e*n*cos(2*d*x^n + 2*c) + d*e*n) + (e*x)^n*a^2/(e*n) - ( 
e^n*log(cos(d*x^n)^2 + 2*cos(d*x^n)*cos(c) + cos(c)^2 + sin(d*x^n)^2 - 2*s 
in(d*x^n)*sin(c) + sin(c)^2) - e^n*log(cos(d*x^n)^2 - 2*cos(d*x^n)*cos(c) 
+ cos(c)^2 + sin(d*x^n)^2 + 2*sin(d*x^n)*sin(c) + sin(c)^2))*a*b/(d*e*n)
 

Giac [F]

\[ \int (e x)^{-1+n} \left (a+b \csc \left (c+d x^n\right )\right )^2 \, dx=\int { {\left (b \csc \left (d x^{n} + c\right ) + a\right )}^{2} \left (e x\right )^{n - 1} \,d x } \] Input:

integrate((e*x)^(-1+n)*(a+b*csc(c+d*x^n))^2,x, algorithm="giac")
 

Output:

integrate((b*csc(d*x^n + c) + a)^2*(e*x)^(n - 1), x)
 

Mupad [B] (verification not implemented)

Time = 17.33 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.28 \[ \int (e x)^{-1+n} \left (a+b \csc \left (c+d x^n\right )\right )^2 \, dx=\frac {a^2\,x\,{\left (e\,x\right )}^{n-1}}{n}-\frac {b^2\,x\,{\left (e\,x\right )}^{n-1}\,2{}\mathrm {i}}{d\,n\,x^n\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x^n\,2{}\mathrm {i}}-1\right )}-\frac {2\,a\,b\,x\,\ln \left (-a\,b\,{\left (e\,x\right )}^{n-1}\,4{}\mathrm {i}-a\,b\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\,{\mathrm {e}}^{d\,x^n\,1{}\mathrm {i}}\,{\left (e\,x\right )}^{n-1}\,4{}\mathrm {i}\right )\,{\left (e\,x\right )}^{n-1}}{d\,n\,x^n}+\frac {2\,a\,b\,x\,\ln \left (a\,b\,{\left (e\,x\right )}^{n-1}\,4{}\mathrm {i}-a\,b\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\,{\mathrm {e}}^{d\,x^n\,1{}\mathrm {i}}\,{\left (e\,x\right )}^{n-1}\,4{}\mathrm {i}\right )\,{\left (e\,x\right )}^{n-1}}{d\,n\,x^n} \] Input:

int((a + b/sin(c + d*x^n))^2*(e*x)^(n - 1),x)
 

Output:

(a^2*x*(e*x)^(n - 1))/n - (b^2*x*(e*x)^(n - 1)*2i)/(d*n*x^n*(exp(c*2i + d* 
x^n*2i) - 1)) - (2*a*b*x*log(- a*b*(e*x)^(n - 1)*4i - a*b*exp(c*1i)*exp(d* 
x^n*1i)*(e*x)^(n - 1)*4i)*(e*x)^(n - 1))/(d*n*x^n) + (2*a*b*x*log(a*b*(e*x 
)^(n - 1)*4i - a*b*exp(c*1i)*exp(d*x^n*1i)*(e*x)^(n - 1)*4i)*(e*x)^(n - 1) 
)/(d*n*x^n)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.96 \[ \int (e x)^{-1+n} \left (a+b \csc \left (c+d x^n\right )\right )^2 \, dx=\frac {e^{n} \left (-\cos \left (x^{n} d +c \right ) b^{2}+x^{n} \sin \left (x^{n} d +c \right ) a^{2} d +2 \,\mathrm {log}\left (\tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right )\right ) \sin \left (x^{n} d +c \right ) a b \right )}{\sin \left (x^{n} d +c \right ) d e n} \] Input:

int((e*x)^(-1+n)*(a+b*csc(c+d*x^n))^2,x)
 

Output:

(e**n*( - cos(x**n*d + c)*b**2 + x**n*sin(x**n*d + c)*a**2*d + 2*log(tan(( 
x**n*d + c)/2))*sin(x**n*d + c)*a*b))/(sin(x**n*d + c)*d*e*n)